 Hello and welcome to the session, I am Shashi and I am going to help you with the following question. Question says, solve the following linear programming problems graphically, maximize z is equal to 3x plus 2y subject to x plus 2y less than equal to 10, 3x plus 5 less than equal to 15, xy greater than equal to 0. Let us now start with the solution, now according to the given problem we have to maximize z is equal to 3x plus 2y subject to constraints x plus 2y less than equal to 10, 3x plus 5 less than equal to 15, x greater than equal to 0 and y greater than equal to 0. Now for drawing the graph and for finding the feasible region, first of all we will draw a line x plus 2y is equal to 10 corresponding to this inequality. Now points 0 comma 5 and 10 comma 0 lie on the line x plus 2y is equal to 10. Now we will plot these two points on the graph and obtain this line by joining these two points. Now clearly we can see this point represents 0 comma 5 and this point represents 10 comma 0 joining these two points we get the line x plus 2y is equal to 10. Now this line divides the plane into two half planes, we will consider the plane satisfying the inequality x plus 2y is less than 10 that is we will consider the half plane containing the origin 0 0. Now we will draw a line 3x plus y is equal to 15 corresponding to this inequality. Now points 0 comma 15 and 5 comma 0 lie on the line 3x plus y is equal to 15. Now we will plot these two points on the same graph and we will draw this line by joining these two points on the same graph. Now clearly we can see this point represents 0 comma 15 and this point represents 5 comma 0 joining these two points we get the line 3x plus y is equal to 15. Now this line divides the plane into two half planes we will consider the plane that satisfies the inequality 3x plus y is less than 15 or we can say we will consider the plane that contains the origin 0 0. Now clearly we can see x is greater than equal to 0 and y is greater than equal to 0 implies that the graph lies in the first quadrant only. Now this shaded portion in the graph which forms a convex polygon is the feasible region satisfying all the given constraints. Let us mark this point as b this point as c and this point as d clearly we can see coordinates of point d are 4 comma 3 these two lines intersect each other at point c and coordinates of point c are 4 3. Now let us recall that the common region determined by all the constraints including the non-negative constraints x greater than equal to 0 y greater than equal to 0 of a linear programming problem is called feasible region for the problem. Now convex polygon ovcd represents the feasible region of the given linear programming problem. Now we can write the shaded portion in the graph that is the convex polygon ovcd is the feasible region satisfying all the given constraints. Now corner points of the feasible region are 0 comma 0 5 comma 0 4 comma 3 and 0 comma 5. Now we know according to corner point method the maximum or minimum value of a linear objective function over a convex polygon occurs at some vertex of the polygon. So we will find the value of z at all the corner points the corner points of the feasible region which has coordinates 0 comma 0 b having coordinates 5 comma 0 c having coordinates 4 comma 3 and d having coordinates 0 comma 5. Now we are given objective function is z is equal to 3 x plus 2 y. Now first of all let us find out the value of z at 0 comma 0 z is equal to 0 at point 0 comma 0. Now we will find the value of z at point 5 comma 0 it is equal to 3 multiplied by 5 plus 2 multiplied by 0 substituting 5 for x and 0 for y in this expression we get z is equal to 15. So we can write z is equal to 15 at point 5 comma 0 similarly we can find the value of z at other corner point 4 comma 3 substituting 4 for x and 3 for y in this objective function we get z is equal to 3 multiplied by 4 plus 2 multiplied by 3 which is further equal to 18. So we get z is equal to 18 at point 4 comma 3. Now we will find the value of z at point 0 comma 5 substituting 0 for x and 5 for y in this expression we get z is equal to 10. So we can write z is equal to 10 at point 0 comma 5. Now clearly we can see maximum value of z is 18 which occurs at point 4 comma 3. So we can write hence maximum value of z is equal to 18 which occurs at point 4 comma 3. So this is our required answer this completes the session hope you understood the solution take care and have a nice day.